Question

Find the roots of the equation by completing the square: 3x^2-6x-2=0. Prove your answer by solving by the quadratic formula.

Answer 1

`3x^2-6x-2=0\\\\(\sqrt3\ x)^2-2\cdot\sqrt3\ x\cdot\sqrt3+(\sqrt3)^2-(\sqrt3)^2-2=0\\\\(\sqrt3\ x-\sqrt3)^2-3-2=0\\\\(\sqrt3\ x-\sqrt3)^2=5\iff\sqrt3\ x-\sqrt3=-\sqrt5\ \vee\ \sqrt3\ x-\sqrt3=\sqrt5\\\\\sqrt3\ x=\sqrt3-\sqrt5\ \vee\ \sqrt3\ x=\sqrt3+\sqrt5\ \ \ \ \ |multiply\ both\ sides\ by\ \sqrt3\\\\3x=3-√(15)\ \vee\ 3x=3+√(15)\ \ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\\x=(3-√(15))/(3)\ \vee\ x=(3+√(15))/(3)`




`Prove:\\\\3x^2-6x-2=0\\a=3;\ b=-6;\ c=-2\\\Delta=b^2-4ac\to\Delta=(-6)^2-4\cdot3\cdot(-2)=36+24=60\\\\x_1=(-b-\sqrt\Delta)/(2a);\ x_2=(-b+\sqrt\Delta)/(2a)\\\\\sqrt\Delta=√(60)=√(4\cdot15)=\sqrt4\cdot√(15)=2√(15)\\\\x_1=(6-2√(15))/(2\cdot3)=(3-√(15))/(3)\ \vee\ x_2=(6+2√(15))/(2\cdot3)=(3+√(15))/(3)`

Answer 2

Calculating delta:
Δ=b²-4ac
a=3
b=-6
c=-2
Δ=36-4*3*(-2)=36+24=60
√Δ=√60
Delta is positive so there are two roots:
x1=
`\frac{ -b+ \sqrt[2]{delta} }{2a}`
x1=
`(6+ √(delta) )/(2*3)`=
`(3+ √(15) )/(3)`
x2=
`(3- √(15) )/(3)`